How is the output of a monostable multivibrator affected if the trigger pin is kept at a negative voltage (0V)? The upper voltage reference comparator will change its output to logic 1, while the lower reference comparator is already at logic 1. As a result the input to the RS flip flop goes into forbidden state (R=1,S=1). What happens at the output?
\$\begingroup\$
\$\endgroup\$
4
-
\$\begingroup\$ it is impossible to guess about the behavior of an unknown circuit \$\endgroup\$– jsotolaCommented Jan 20, 2019 at 19:26
-
\$\begingroup\$ This sounds a little like homework. \$\endgroup\$– deathismyfriendCommented Jan 20, 2019 at 20:02
-
1\$\begingroup\$ 0V is not a negative voltage. \$\endgroup\$– FinbarrCommented Jan 20, 2019 at 23:51
-
1\$\begingroup\$ What happens is that one input dominates. \$\endgroup\$– Spehro 'speff' PefhanyCommented Feb 25, 2020 at 22:16
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
The 555 device if connected in a standard monostable mode will generate an output pulse at the falling edge of the trigger pin signal. If the trigger pulse signal goes back high before the normal ending of the output pulse then there is no effect on the output pulse. If the trigger pulse stays low longer than the normal timing of the output pulse then the output pulse will stay high until such time that the trigger input goes back low. These comments apply to a circuit wired up as follows:
If you want to have the trigger pulse level not interfere with the output pulse then you have to make sure that the trigger pulse is always shorter than the nominal output pulse width. One way to do this is to use an R/C coupling of the trigger input so that it is just a short spike just enough to trigger the 555. Here is a circuit configuration that is setup that way.
answered Jan 20, 2019 at 19:47